Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A. n (n+1) (n-4) = 7*8*3 and if n = 8 --> 8*9*4
B. n (n+2) (n-1) = 7*9*6 and if n = 8 --> 8*10*7
C. n (n+3) (n-5) = 7*10*5; eliminate as there are no multiples of 3
D. n (n+4) (n-2) = 7*11*5; eliminate as there are no multiples of 3
E. n (n+5) (n-6) = 7*12*1 and and if n = 8 --> 8*13*2

If n is an integer greater than 6, which of the following must be divisible by 3? A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

anything in the form of (n-1) (n) (n+1) is divvisible by 3. in other word, a product of any 3 consecutie intevers is divisible by 3.

A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)
B. n (n+2) (n-1) is equivalant to (n+1) missing.
C. n (n+3) (n-5) is equivalant to (n-1) missing and n repeating.
D. n (n+4) (n-2) is equivalant to odd/even consqcutive integers
E. n (n+5) (n-6) is equivalant to (n+1) missing and n repeating.

If n is an integer greater than 6, which of the following must be divisible by 3?

A) N(N+1)(N-4) B) N(N+2)(N-1) C) N(N+3)(N-5) D) N(N+4)(N-2) E) N(N+5)(N-6)

Is there an easy way to solve this?

Substituting 3 consecutive numbers might be the easiest way to solve this question as shown above.

Algebraic solution:

The product of 3 numbers to be divisible by 3 at least one of them must be divisible by 3. So, to ensure that the product of 3 integers shown is divisible by 3 all 3 numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3. We should have something like \(n(n+1)(n+2)\) (for example: if n divided by 3 yields remainder of 1, then n+1 yields remainder of 2 and n+2 yields remainder of 0, thus it's divisible by 3 OR if n divided by 3 yields remainder of 2, then n+2 yields remainder of 1 and n+1 yields remainder of 0, thus it's divisible by 3).

Only option A satisfies this, because \(n(n+1)(n-4)=n(n+1)(n-6+2)\) and \(n-6\) has the same remainder as \(n\) upon division by 3 thus we can replace it by \(n\).

Re: If n is an integer greater than 6, which of the following must be divi [#permalink]

Show Tags

13 Sep 2010, 13:16

2

This post received KUDOS

Same concept different way of thinking

How do I ensure that a product of three numbers is divisible by 3 ? Step 1 : I divide the set of integers into three parts such that one of the parts contains exactly and only the multiples of 3 Step 2 : I ensure that I chose my numbers such that each belongs to a different part Step 3 : If I achieve this, I must have exactly one number which belongs to the set of multiples of three. Hence their product is divisible by three.

How do I achieve this ? Step 1 : The division I choose is numbers of the form \(3\alpha, 3\alpha+1, 3\alpha+2\). The union of these sets is all integers and they have no overlap between them, also one of them contains all and only multiples of 3. Step 2 : Simplest way to choose three numbers such that exactly one belongs to each of the above sets is 3 consecutive numbers. This choice will always satisfy what I am looking for. Hence n*(n-1)*(n+1)

Important Note : If I shift any number by 3, its constituent set amongst \(\alpha, 3\alpha+1, 3\alpha+2\) does not change

So how do we solve this question ? Start from the expression n*(n-1)*(n+1) and see if by adding or subtracting 3 or multiples of 3 from one or more of the three terms we can get any target expression.

A) Satisfies B) n-1 repeats twice C) n repeats twice D) n+1 repeats twice E) n repeats twice

Re: If n is an integer greater than 6, which of the following must be divi [#permalink]

Show Tags

13 Sep 2010, 17:39

I have seen many questions that derive upon n n-1 n+1 being divisible by 3.. Does this and the logic below.. can both be extended to any sequence? Meaning 4..5..6.. consecutive numbers? It would seem that is the case?

Can you explain what do you mean by n-1 repeats twice in B? A) Satisfies B) n-1 repeats twice C) n repeats twice D) n+1 repeats twice E) n repeats twice

shrouded1 wrote:

Same concept different way of thinking

How do I ensure that a product of three numbers is divisible by 3 ? Step 1 : I divide the set of integers into three parts such that one of the parts contains exactly and only the multiples of 3 Step 2 : I ensure that I chose my numbers such that each belongs to a different part Step 3 : If I achieve this, I must have exactly one number which belongs to the set of multiples of three. Hence their product is divisible by three.

How do I achieve this ? Step 1 : The division I choose is numbers of the form \(3\alpha, 3\alpha+1, 3\alpha+2\). The union of these sets is all integers and they have no overlap between them, also one of them contains all and only multiples of 3. Step 2 : Simplest way to choose three numbers such that exactly one belongs to each of the above sets is 3 consecutive numbers. This choice will always satisfy what I am looking for. Hence n*(n-1)*(n+1)

Important Note : If I shift any number by 3, its constituent set amongst \(\alpha, 3\alpha+1, 3\alpha+2\) does not change

So how do we solve this question ? Start from the expression n*(n-1)*(n+1) and see if by adding or subtracting 3 or multiples of 3 from one or more of the three terms we can get any target expression.

A) Satisfies B) n-1 repeats twice C) n repeats twice D) n+1 repeats twice E) n repeats twice

Re: If n is an integer greater than 6, which of the following must be divi [#permalink]

Show Tags

14 Sep 2010, 02:00

mainhoon wrote:

I have seen many questions that derive upon n n-1 n+1 being divisible by 3.. Does this and the logic below.. can both be extended to any sequence? Meaning 4..5..6.. consecutive numbers? It would seem that is the case?

Can you explain what do you mean by n-1 repeats twice in B? A) Satisfies B) n-1 repeats twice C) n repeats twice D) n+1 repeats twice E) n repeats twice

The important point to note is that adding or subtracting a multiple of 3 does not change the constituent set And that we need to ensure there is one element from each set, \(3\alpha, 3\alpha+1,3\alpha+2\)

So B for instance is :

\(n*(n+2)*(n-1) \equiv n*(n+2-3)*(n-1) \equiv n*(n-1)*(n-1)\) (I dont mean to say these expressions are equal, just equal vis-a-vis what their constituent sets are)

So the number of type (n-1) repeats but there is none of type (n+1) hence there is no guarantee this product is divisible by 3. And the counter example would be a case where n+1 is divisible by 3. Eg. n=11, expression = 11*13*10, not divisible by 3.
_________________

Re: If n is an integer greater than 6, which of the following must be divi [#permalink]

Show Tags

14 Sep 2010, 02:07

1

This post received KUDOS

1

This post was BOOKMARKED

The best approach is to back-solve, but to learn the concept read the reply of Bunnel. That concept is very important.

If you have a number n , highest number of remainders are 0 to n-1

Since n =3, the remainders are 0,1,2. Since the remainders are consecutive numbers, we should consider 3 consecutive numbers to back solve.
_________________

If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+1)(n-4) b) n(n+2)(n-1) c) n(n+3)(n-5) d) n(n+4)(n-2) e) n(n+5)(n-6)

I chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

As for your doubt: n(n+2)(n-1) is NOT a product of 3 consecutive integers. If it were: n(n+1)(n-1) or n(n+2)(n+1) then YES, but in its current form it's not.
_________________

If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+1)(n-4) b) n(n+2)(n-1) c) n(n+3)(n-5) d) n(n+4)(n-2) e) n(n+5)(n-6)

I chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

b) n(n+2)(n-1) can you justify these are consecutive integers? No they are not.

Now take n = 3k , n = 3k+1 n = 3k+2..put in all the choices. If by putting all the values of n we get it is divisible by 3, then it is correct answer choice.

A is correct. It will hardy take 10 sec per choice as we have to consider only 3k+1 and 3k+2.
_________________

If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+1)(n-4) b) n(n+2)(n-1) c) n(n+3)(n-5) d) n(n+4)(n-2) e) n(n+5)(n-6)

I chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

As for the rule you quote it's correct:

• If \(k\) is odd, the sum of \(k\) consecutive integers is always divisible by \(k\). Given \(\{9,10,11\}\), we have \(k=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If \(k\) is even, the sum of \(k\) consecutive integers is never divisible by \(k\). Given \(\{9,10,11,12\}\), we have \(k=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. Given \(k=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.
_________________

If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+1)(n-4) b) n(n+2)(n-1) c) n(n+3)(n-5) d) n(n+4)(n-2) e) n(n+5)(n-6)

I chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

Well, B is not 3 consecutive integers. That's why your answer is wrong.

The trick is that if you divide integers into 3 sets 3K, 3K+1, 3K+2 and figure out a strategy such that 3 numbers you pick are one each from these sets, then the product has to be divisble be three. The easiest way to do this is selecting 3 consecutive numbers

So start with n(n-1)(n+1) which are consecutive. Now adding or subtracting three from any of these numbers doesnt change its source set. so thats how you figure out the answer.

a) n(n+1)(n-4) ---> n(n+1)(n-4+3) --> This is it ! b) n(n+2)(n-1) --> n(n+2-3)(n-1) --> n-1 repeats c) n(n+3)(n-5) --> n(n+3-3)(n-5+6) --> n repeats d) n(n+4)(n-2) --> n(n+4-3)(n-2+3) --> n+1 repeats e) n(n+5)(n-6) --> n(n+5-6)(n-6+6) --> n repeats
_________________

WE: Information Technology (Internet and New Media)

Re: If n is an integer greater than 6, which of the following must be divi [#permalink]

Show Tags

06 Oct 2010, 16:10

thanks for all the replies! i cant believe how silly i was... my answer wasnt a consec. int. sequence at all! for some reason i misread it and thought it was like an unsorted consec. int. problem like 7,6,8. if it was n + 1 instead of n+2 then it would have been a consec. int. problem... just goes to show dont study when your tired! but thanks for the answers cos i didnt know how to solve it anyway and the og answer wasnt that great.

Re: If n is an integer greater than 6, which of the following must be divi [#permalink]

Show Tags

07 Oct 2010, 13:06

The condition of divisibility by 3 is that the sum of the digits of the number must be divisible by 3.

I mean in case of A, 3N-3 should be divisible by 3 and it is ,so A is the answer. In remaining options all are not divisible by 3 as they are not being subtracted or added by the multiples of 3 from 3N i.e (3N+1,3N-2,3N+2,3N-1)

Re: If n is an integer greater than 6, which of the following [#permalink]

Show Tags

23 Mar 2012, 21:04

IMO A.

I used an arbitrary number greater than 6 and then filled each equation out. if you happen to have chosen a number that makes more than 1 answer correct, choose a different number and check the ones that were previously correct.
_________________

If n is an integer greater than 6, which of the following must be divisible by 3?

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.

For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.